Theorem REF  improves the corresponding upper bound in Theorem 1.9 in [1]} by a factor of \(n^{-1/3}\)  (up to some constants). The latter follows from
\(\frac{K_nd^2}{ n}=\frac{ (K_n^{1/2}d /n^{1/3})^2}{n^{1/3}}\quad \text{and}\quad {\mathrm {tr}}(A_pA_p^\top )\leqslant \Vert A_pA_p^\top \Vert  {\mathrm {tr}}(I_p)=p\Vert A_p\Vert ^2 .\) 
